Here's a parameterization of a sphere, for $0 < \theta < 2\pi$ and $0 < \varphi < \pi$ : $\vec{v}(\theta, \varphi) = (r\cos(\theta)\sin(\varphi), r\sin(\theta)\sin(\varphi), r\cos(\varphi))$ What is the inward-pointing vector normal to the area element of this sphere given $r = 4$, $\theta = \dfrac{\pi}{6}$, and $\varphi = \dfrac{3\pi}{4}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\left( -4\sqrt{3}, -4, 8 \right)$ (Choice B) B $\left( -4\sqrt{3}, 8, -4 \right)$ (Choice C) C $\left( 4\sqrt{3}, -8, 4 \right)$ (Choice D) D $\left( 4\sqrt{3}, 4, -8 \right)$
The vector normal to the area element describes what's perpendicular to the surface, and the vector's magnitude represents the area of a tiny rectangle along the parameterization. After we compute it, we need to check whether it's pointing inwards or outwards. $\text{normal to area element} = \dfrac{\partial \vec{v}}{\partial \theta} \times \dfrac{\partial \vec{v}}{\partial \varphi}$ Let's calculate the cross product. $\begin{aligned} \dfrac{\partial \vec{v}}{\partial \theta} \times \dfrac{\partial \vec{v}}{\partial \varphi} &= \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \\ -r\sin(\theta)\sin(\varphi) & r\cos(\theta)\sin(\varphi) & 0 \\ \\ r\cos(\theta)\cos(\varphi) & r\sin(\theta)\cos(\varphi) & -r\sin(\varphi) \end{pmatrix} \\ \\ &= \left( -r^2 \cos(\theta) \sin^2(\varphi) - 0 \right) \hat{\imath} + \left(-r^2\sin(\theta)\sin^2(\varphi) - 0 \right) \hat{\jmath} \\ \\ &+ \left(-r^2\sin^2(\theta)\cos(\varphi)\sin(\varphi) - r^2\cos^2(\theta)\cos(\varphi)\sin(\varphi \right) \hat{k} \\ \\ &= \left( -r^2\cos(\theta)\sin^2(\varphi) \right) \hat{\imath} + \left( -r^2\sin(\theta)\sin^2(\varphi) \right) \hat{\jmath} \\ \\ &+ \left(-r^2\cos(\varphi)\sin(\varphi) \right) \hat{k} \end{aligned}$ Plugging in $r = 4$, $\theta = \dfrac{\pi}{6}$, and $\varphi = \dfrac{3\pi}{4}$, we get the vector normal to the area element: $\begin{aligned} \dfrac{\partial \vec{v}}{\partial \theta} \times \dfrac{\partial \vec{v}}{\partial \varphi} &= \left( -(4)^2\cos\left(\dfrac{\pi}{6}\right)\sin^2\left( \dfrac{3\pi}{4} \right) \right) \hat{\imath} \\ \\ & + \left( -(4)^2\sin\left(\dfrac{\pi}{6}\right)\sin^2\left( \dfrac{3\pi}{4} \right) \right) \hat{\jmath} \\ \\ &+ \left(-(4)^2\cos\left( \dfrac{3\pi}{4} \right) \sin\left( \dfrac{3\pi}{4} \right) \right) \hat{k} \\ \\ &= \left(-4\sqrt{3}\right) \hat{\imath} + \left(-4\right) \hat{\jmath} + \left(-8\right) \hat{k} \\ \\ &= \left( -4\sqrt{3}, -4, 8 \right) \end{aligned}$ The final step is to check whether this points inwards or outwards. We know that a sphere's outward facing normal vector points directly away from the origin. Because $\theta = \dfrac{\pi}{6}$ and $\varphi = \dfrac{3\pi}{4}$ correspond to a positive $x$, positive $y$, and negative $z$, the inward normal vector must have a negative $x$, negative $y$, and positive $z$. Our calculation matches this inward normal vector. Therefore, the inward-pointing vector normal to the area element is $\left( -4\sqrt{3}, -4, 8 \right)$.